Tuesday

July 29, 2014

July 29, 2014

Posted by **katarina** on Tuesday, May 15, 2012 at 7:40pm.

2. Find the area of the region bounded by x=y^2+6, x=0 , y=-6, and y=7. Show all work required in #1.

3. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about y=3. Show the integral and give an exact answer.

- calculus -
**Reiny**, Tuesday, May 15, 2012 at 9:44pmThere are 3 questions here.

You show no work , nor do you tell us where your difficulty is.

I will start you off with #1

we need their intersection points

x^2 + 6x + 9 = 5x + 15

x^2 + x - 6 = 0

(x+3)(x-2) = 0

x = -3 or x = 2

if x = -3, y = 0

if x = 2 , y = 25

from x = -3 to x = 2, g(x) > f(x), so the effective height

= 5x + 15 - x^2 - 6x - 9

= 6 - x - x^2

Area = ∫(6 - x - x^2) dx from x = -3 to 2

= [6x - (1/2)x^2 - (1/3)x^3] from -3 to 2

= (12 - (1/2)(4) - (1/3)(8) ) - (-18 - (1/2)(9) - (1/3)(-27) )

= 12 - 2 - 8/3 + 18 - 9/2 + 9

= 179/6

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